2 Marks Questions

Question 12 Marks

Two electric bulbs P and Q have their resistances in the ratio of 1 : 2. They are connected in series across a battery. Find the ratio of the power dissipation of these bulbs.

Answer

For series combination, power dissipated by a bulb is directly proportional to its resistance.
$ P \propto R $
$ \frac{P_1}{P_2} = \frac{R_1}{R_2} = \frac{1}{2} $

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Question 22 Marks

Write down two uses of potentiometer.

Answer

Potentiometer is an accurate instrument to measure the electromotive force of a cell or the potential difference between two points of an electrical circuit.
In fact, the potentiometer behaves like a voltmeter whose own resistance is infinite.

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Question 32 Marks

Define Volume density of charge. Write its S.I. unit.

Answer

Volume charge density $ (\rho) $ is the quantity of charge per unit volume, measured in the S.I. system in coulombs per cubic meter (cm^-3), at any point in a volume.

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Question 42 Marks

Four $ 12~\Omega $ resistances are connected in parallel. There such combinations are connected in series. What will be the total resistance ?

Answer

Four $ 12~\Omega $ resistances are connected (i) Parallel.
$ \Rightarrow \frac{1}{R_P} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} $
$ R_P = 3~\Omega $
Now, $ 3~\Omega $ resistances (3 in number) are connected in series.
$ \Rightarrow R_S = 3 + 3 + 3 = 9~\Omega $.

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Question 52 Marks

A bulb rated 100 W-220 V is connected across 240 volt. Calculate electric power loss.

Answer

Resistance of the bulb (rated: 100 W-220 V) is given by
$ R = \frac{V^2}{P} = (\frac{220 \times 220}{100})~\Omega = 484~\Omega $
Now power loss when the bulb is connected across
$ V' = 240 $ volt;
$ P' = \frac{V'^2}{R} = (\frac{240 \times 240}{484})~\Omega = 119~W $

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Question 62 Marks

A wire of resistance 16 ohm is clongated to double its lengths find the new resistance ?

Answer

When a wire is streteched to double its length, its area of cross-section becomes half of the initial value
For First Case $R=\rho 1 / A$
For Second Case $R^{\prime}=\rho 1 / A=>\rho(21) /(A / 2)$
= 4R
$=4 \times 16=64$ ohm

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Question 72 Marks

Define electric field intensity and equipotential surface in current carrying conductor.

Answer

Electric Field Intensity : In a current carrying conductor the ratio of potential difference applied across the ends of the conductor and the length of the conductor is equal to the electric field intensity at any point in the conductor. i.e.
$ E = V/l $
where $ V = $ potential difference and $ l = $ length.
Equipotential Surface : In a currnet carrying conductor electric currnet flows from the end of higher potential towards the end of lower potential. Hence electric potential decreases with increase the distance from the end of higher potential toward the end of lower potential but the electric potential at all points lying on any cross-section of the conductor is the same. Hence any cross-section of the conductor is equipotential surface.

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Question 82 Marks

Write the expression for the resistivity of the conductor and explain each term of the expression.

Answer

Resistivity : $ \rho = \frac{E}{j} $; where $ E = $ Intensity of electric field in the conductor and $ j = $ current density in the conductor.
It may also be written as follows also :
$ \rho = (\frac{R \times A}{L}) $; where $ R = $ Resistance of the conductor,
$ A = $ area of cross-section of the conductor and $ L = $ length of the conductor.

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Question 92 Marks

What do you mean by drift velocity and mobility ?

Answer

Drift Velocity : It is defined as the small constant velocity with which free electrons get drifted towards the positive end of the conductor under the influence of the electric field developed within the conductor by applying potential difference across the ends of the conductor. It is of the order of $ 10^{-4} m/s $.
Mobility : Mobility of a charge carrier is defined as the drift velocity per unit electric field strength i.e.,
$ \mu = v_d / E $.
Its S.I. unit is $ m^2/Vs $.

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Question 102 Marks

Two resistances are connected in parralel in the ratio of 3: 4. Compare the results of the heat generated in them.

Answer

$ \frac{R_1}{R_2} = \frac{3}{4} $
For equal potentiometer
$ \frac{H_1}{H_2} = \frac{R_2}{R_1} = \frac{4}{3} $
$ H_1 : H_2 = 4:3 $

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Question 112 Marks

What are ohmic and non-ohmic resistances? Write down one example of both.

Answer

Ohmic devices are the devices that follow Ohm's law.
Examples : wire and resistor.
Non-Ohmic devices are devices that do not follow Ohm's law.
Examples : vacuum tubes and thermistors.

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